Question

A wheel initially rotating at 12 rad/s decelerates uniformly to rest in 0.4 s. If the wheel has a rotational inertia of 0.5 kg.m², what is the magnitude of the torque causing this deceleration? (A) 1.5 N.m (B) 15 N.m (C) 30 Nm (D) 38 Nm​

Answer 1

(B) 15 N.m

Step-by-step explanation:

The deceleration of the wheel is first found by using the following formula:

`\alpha = (\omega_f - \omega_i)/(t)`

where,

α = angular acceleration = ?

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = 12 rad/s

t = time = 0.4 s

Therefore,

`\alpha = (0\ rad/s - 12\ rad/s)/(0.4\ s)\\\\\alpha = -30\ rad/s^2`

here, the negative sign shows deceleration.

Now, we find the torque responsible for this deceleration:

`\tau = I\alpha`

where,

τ = torque = ?

I = rotational inertia = 0.5 kg.m²

Therefore,

`\tau = (0.5\ kg.m^2)(30\ rad/s^2)\\`

τ = 15 N.m

Therefore, the correct answer is:

(B) 15 N.m